Question

A monatomic ideal gas undergoes isothermal expansion from 0.08 m3 to 0.45 m3 at a constant temperature (initial pressure is 430 kPa). What are its (a) internal energy change ( Δ E ), (b) net heat transfer ( Q ), and (c) net work done ( W )? Use negative quantity for heat transfer out of the system or work done on the system

Answer 1

Answer :

(a) Internal energy change is, zero.

(b) Net heat transfer is, 59416.27 J

(c) Net work done is, -59416.27 J

Explanation : Given,

Initial volume of gas = $0.08m^3=80L$

Final volume of the gas = $0.45m^3=45L$

Initial pressure = 430 kPa

Moles of gas (n) = 1

First we have to calculate the temperature of the gas.

$P_1V_1=nRT$

where,

$P_1$ = initial pressure of gas

$V_1$ = initial volume of gas

n = moles of gas

R = gas constant = 8.314 kPa.L/K.mole

T = temperature of gas

Now put all the given values in the above formula, we get:

$P_1V_1=nRT$

$(430kPa)\times (80L)=(1mole)\times (8.314kPa.L/K.mole)\times T$

$T=4137.59K$

According to the question, this is the case of isothermal reversible expansion of gas.

As per first law of thermodynamic,

$\Delta U=q+w$

where,

$\Delta U$ = internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

$\Delta U=0$

$q=-w$

The expression used for work done will be,

$w=-nRT\ln (\frac{V_2}{V_1})$

where,

w = work done on the system = ?

n = number of moles of gas  = 1

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 4137.59 K

$V_1$ = initial volume of gas  = $0.08m^3$

$V_2$ = final volume of gas  = $0.45m^3$

Now put all the given values in the above formula, we get :

$w=-1mole\times 8.314J/moleK\times 4137.59K\times \ln (\frac{0.45m^3}{0.08m^3})$

$w=-59416.27J$

Net work done is, -59416.27 J

and,

q = - w = -(-59416.27 J) = 59416.27 J

Net heat transfer is, 59416.27 J